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This is a simple question, but I just don't understand how we determine the poles and zeros of a rational system function.

For example, for the LTI system described by this constant coefficient difference equation $$ y[n]-\frac{5}{2}y[n-1]+y[n-2]=x[n] $$ we can determine that

$$ H(z)=\frac{1}{(1-\frac{1}{2}z^{-1})(1-2z^{-1})} $$

I understand why there are poles at $z=\frac{1}{2}$ and $z=2$, but I don't understand why there are two zeros at $z=0$. Even if I multiply $H(z)$ through by $z$, there would only be one zero, correct?

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Just multiply numerator and denominator by $z^2$ to obtain

$$H(z)=\frac{z^2}{(z-\frac12)(z-2)}$$

from which you see that there's a double zero at $z=0$. Note that the number of zeros and poles is always equal if you include poles and zeros at infinity.

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  • $\begingroup$ If I multiplied the numerator and denominator by z, then wouldn't I end up with z/((z-1/2)(z-2))? Where does the z^2 come from? $\endgroup$ – user50420 Sep 30 '18 at 17:16
  • $\begingroup$ @user50420: That was a typo, I meant "by $z^2$". $\endgroup$ – Matt L. Sep 30 '18 at 17:21
  • $\begingroup$ I see what you mean now. This was a trivial question, but for some reason I forgot how multiplication works. Thank for your help. $\endgroup$ – user50420 Sep 30 '18 at 17:29

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